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Sagot :
Let's break down the problem step by step and identify the correct vectors:
1. Ship's speed and direction:
- The ship is moving at \(30 \, \text{miles/hour}\) at an angle of \(30^\circ\) south of east.
2. Translate the ship's direction to standard angle measurement:
- In this case, standard angle measurement is counter-clockwise from the positive \(x\)-axis (east direction).
- Since the ship's direction is \(30^\circ\) south of east, we convert this to \(330^\circ\) (i.e., \(360^\circ - 30^\circ\)).
3. Convert the ship's direction to vector components:
- The angle \(\theta_s\) corresponding to \(330^\circ\) is used to find the components:
[tex]\[ \text{Ship's vector} = \left(30 \cos 330^\circ, 30 \sin 330^\circ\right) = \left(30 \cos (-30^\circ), 30 \sin (-30^\circ)\right) \][/tex]
- Calculating the components:
[tex]\[ \left(30 \times \frac{\sqrt{3}}{2}, 30 \times -\frac{1}{2}\right) = \left(15\sqrt{3}, -15\right) \approx \left(25.98, -15\right) \][/tex]
4. Water current's speed and direction:
- The water current is moving at \(5 \, \text{miles/hour}\) at an angle of \(20^\circ\) east of north.
5. Translate the water current's direction to standard angle measurement:
- The water's direction is \(20^\circ\) east of north, which directly translates to \(70^\circ\) (from \(90^\circ\) - \(20^\circ\)) in standard angle measurement.
6. Convert the water current's direction to vector components:
- The angle \(\theta_w\) corresponding to \(70^\circ\) is used to find the components:
[tex]\[ \text{Water current's vector} = \left(5 \cos 70^\circ, 5 \sin 70^\circ\right) \][/tex]
- Calculating the components:
[tex]\[ (5 \times \cos 70^\circ, 5 \times \sin 70^\circ) \approx (5 \times 0.342, 5 \times 0.94) \approx (1.71, 4.7) \][/tex]
7. Resultant vector:
- The resultant vector of the ship's actual motion is found by summing the ship's vector and the water current's vector:
- Combining:
[tex]\[ \left(25.98 + 1.71, -15 + 4.7\right) = \left(27.69, -10.3\right) \][/tex]
8. Identifying correct choices:
- From the possible choices given, we match them to our calculations:
- Ship's vector: \(\langle 25.98, -15 \rangle\)
- Water's vector: \(\langle 1.71, 4.7 \rangle\)
- Resultant vector: \(\langle 27.69, -10.3 \rangle\)
Therefore, the correct vectors from the options given are:
- Ship's vector: \(\langle 25.98, -15 \rangle\)
- Water's vector: [tex]\(\langle 1.71, 4.7 \rangle\)[/tex]
1. Ship's speed and direction:
- The ship is moving at \(30 \, \text{miles/hour}\) at an angle of \(30^\circ\) south of east.
2. Translate the ship's direction to standard angle measurement:
- In this case, standard angle measurement is counter-clockwise from the positive \(x\)-axis (east direction).
- Since the ship's direction is \(30^\circ\) south of east, we convert this to \(330^\circ\) (i.e., \(360^\circ - 30^\circ\)).
3. Convert the ship's direction to vector components:
- The angle \(\theta_s\) corresponding to \(330^\circ\) is used to find the components:
[tex]\[ \text{Ship's vector} = \left(30 \cos 330^\circ, 30 \sin 330^\circ\right) = \left(30 \cos (-30^\circ), 30 \sin (-30^\circ)\right) \][/tex]
- Calculating the components:
[tex]\[ \left(30 \times \frac{\sqrt{3}}{2}, 30 \times -\frac{1}{2}\right) = \left(15\sqrt{3}, -15\right) \approx \left(25.98, -15\right) \][/tex]
4. Water current's speed and direction:
- The water current is moving at \(5 \, \text{miles/hour}\) at an angle of \(20^\circ\) east of north.
5. Translate the water current's direction to standard angle measurement:
- The water's direction is \(20^\circ\) east of north, which directly translates to \(70^\circ\) (from \(90^\circ\) - \(20^\circ\)) in standard angle measurement.
6. Convert the water current's direction to vector components:
- The angle \(\theta_w\) corresponding to \(70^\circ\) is used to find the components:
[tex]\[ \text{Water current's vector} = \left(5 \cos 70^\circ, 5 \sin 70^\circ\right) \][/tex]
- Calculating the components:
[tex]\[ (5 \times \cos 70^\circ, 5 \times \sin 70^\circ) \approx (5 \times 0.342, 5 \times 0.94) \approx (1.71, 4.7) \][/tex]
7. Resultant vector:
- The resultant vector of the ship's actual motion is found by summing the ship's vector and the water current's vector:
- Combining:
[tex]\[ \left(25.98 + 1.71, -15 + 4.7\right) = \left(27.69, -10.3\right) \][/tex]
8. Identifying correct choices:
- From the possible choices given, we match them to our calculations:
- Ship's vector: \(\langle 25.98, -15 \rangle\)
- Water's vector: \(\langle 1.71, 4.7 \rangle\)
- Resultant vector: \(\langle 27.69, -10.3 \rangle\)
Therefore, the correct vectors from the options given are:
- Ship's vector: \(\langle 25.98, -15 \rangle\)
- Water's vector: [tex]\(\langle 1.71, 4.7 \rangle\)[/tex]
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